Gronwall inequality for $Z_t \leq \int_0^t k(Z_s)\, ds$

Question

In Ikeda and Watanabe, page 170 one reads

I can't see why the relation $$Z_t \leq \int_0^t \kappa (Z_s)\, ds$$ implies that $Z_t = 0$. How do we use the condition $$\int_{0+} \frac{1}{\kappa(u)}\, du = \infty\tag{*}$$ to get to the conclusion?

Answer 1

Under the extra assumptions that $\color{red}{\kappa(u)>0\text{ for }u>0,\text{ continuous and increasing}}$, we can prove that with the condition $(*)$ we have $$ z(t)\le\int_0^t\kappa(z(s))\,ds\qquad\Rightarrow\qquad z(t)\le 0,\ \forall t\ge 0. $$ We adapt the classical Grönwall's method from linear integral equations. Define $$ u(t)=\int_0^t\kappa(z(s))\,ds. $$ Then by the inequality $z(t)\le u(t)$ and monotonicity of $\kappa$ we get $$ \dot u(t)=\kappa(z(t))\le\kappa(u(t))\quad\Rightarrow\quad\frac{\dot u(t)}{\kappa(u(t))}\le 1. $$ We have $u(0)=0$ and increasing since $\kappa\ge 0$, so $u(t)\ge 0$. Integrating from $0$ to $t$ gives $$ \int_0^{u(t)}\frac{du}{\kappa(u)}\le t<+\infty. $$ If $u(t_0)>0$ for some $t_0>0$ we get contradiction of the above inequality with $(*)$, therefore $u(t)\equiv 0$ and, hence, $z(t)\le 0$, $\forall t\ge 0$.

P.S. In your case, $z(t)\ge 0$, so $z(t)\equiv 0$.